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A Converging Lens of Focal Length 12 Cm and a Diverging Mirror of Focal Length 7.5 Cm Are Placed 5.0 Cm Apart with Their Principal Axes Coinciding - Physics

Sum

A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed 5.0 cm apart with their principal axes coinciding. Where should an object be placed so that its image falls on itself?

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Solution

Let the object be placed at a distance cm from the lens (away from the mirror).
For the convex lens (1st refraction) u = − xf = − 12 cm

From the lens formula:
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\Rightarrow\frac{1}{v}=\frac{1}{( - 12)}+\frac{1}{( - x)}\Rightarrow v=-\left( \frac{12x}{x + 12} \right)\]
Thus, the virtual image due to the first refraction lies on the same side as that of object A'B'.
This image becomes the object for the convex mirror,
For the mirror,
\[u = - \left( 5 + \frac{12x}{x + 12} \right)\]
\[ = - \left( \frac{17x + 60}{x + 12} \right)\]
\[f = - 7 . 5 \text{ cm }\]
From mirror equation,
\[\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\]
\[ \Rightarrow \frac{1}{v} = \frac{1}{- 7 . 5} + \frac{x + 12}{17x + 60}\]
\[ \Rightarrow \frac{1}{v} = \frac{17x + 60 - 7 . 5}{7 . 5(17x + 60)}\]
\[\Rightarrow v = \frac{7 . 5(17x + 60)}{52 . 5 - 127 . 5x}\]
\[ \Rightarrow v = \frac{250(x + 4)}{15x - 100}\]
\[ \Rightarrow v = \frac{50(x + 4)}{(3x - 20)}\]
Thus, this image is formed towards the left of the mirror.
Again for second refraction in concave lens,
\[u = - \left[ \frac{5 - 50(x + 4)}{3x - 20} \right]\]
(assuming that the image of mirror formed between the lens and mirror is 3x − 20),
v = + x (since, the final image is produced on the object A"B")
Using lens formula: 
\[\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\]
\[ \Rightarrow \frac{1}{x}+\frac{1}{\frac{\left[ 5 - 50 (x \times 4) \right]}{3x - 20}}=\frac{1}{- 20}\]
⇒ 25x2 − 1400x − 6000 = 0
⇒          x2 − 56x − 240 = 0
⇒          (x − 60) (x + 4) = 0
Thus,                             x = 60 m
The object should be placed at a distance 60 cm from the lens farther away from the mirror, so that the final image is formed on itself.

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 18 Geometrical Optics
Q 62 | Page 416
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